SOLVE :-
(3x +2y - 7)^2 +(5x + 7y - 13)^2 = 0, for real values of x and y
well, I'm not bad at maths, it's that this question is too challenging, so please HELP HELP HELP !!!!!
SOLVE :-
(3x +2y - 7)^2 +(5x + 7y - 13)^2 = 0, for real values of x and y
well, I'm not bad at maths, it's that this question is too challenging, so please HELP HELP HELP !!!!!
Since squares are always greater than or equal to zero, this equation will only have a solution when each of the squares are equal to zero.
So we need to solve the system.
$\displaystyle 3x+2y-7=0$
$\displaystyle 5x+7y-13=0$
Try either elimination or substition
I got
$\displaystyle x=\frac{23}{11} \mbox{ and }y=\frac{4}{11}$